6 circs in a hex

$BG=DE=2$ . $BD=1$ . $\triangle BCD$ is a 30-60-90 so $CD=EF=\sqrt{3}/3$ .

So one side of the hexagon $CD+DE+EF=CF=2+2\sqrt{3}/3 \approx \boxed{3.1547}$

By symmetry, the radii of the $6$ unit circles can be joined to create a smaller regular hexagon with sides of $2$ .

The height $h$ of a regular hexagon with side $s$ is $h = \sqrt{3}s$ , so the height of the smaller regular hexagon is $2\sqrt{3}$ .

Through proportions of the sides and heights of the two similar regular hexagons we have $\frac{x}{2} = \frac{2 + 2\sqrt{3}}{2\sqrt{3}}$ which solves to $x \approx \boxed{3.154700538}$ .

$2\left(1+\dfrac{1}{\sqrt{3}}\right)=3.154700538...$